New Restrictions on Possible Orders of Circulant Hadamard Matrices

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1 New Restrictions on Possible Orders of Circulant Hadamard Matrices Ka Hin Leung Department of Mathematics National University of Singapore Kent Ridge, Singapore Republic of Singapore Bernhard Schmidt Division of Mathematical Sciences School of Physical & Mathematical Sciences Nanyang Technological University Singapore Republic of Singapore Abstract We obtain several new number theoretic results which improve the field descent method. We use these results to rule out many of the known open cases of the circulant Hadamard matrix conjecture. In particular, the only known open case of the Barker sequence conjecture is settled. 1

2 1 Introduction A circulant Hadamard matrix of order v is a square matrix of the form H = a 1 a 2 a v a v a 1 a v 1 a 2 a 3 a 1 with a i { 1, 1} for all i and HH T = vi. No circulant Hadamard matrix of order larger than 4 has ever been found. This led Ryser [12, p. 134] to the following. Conjecture 1.1 No circulant Hadamard matrix of order larger than 4 exists. Conjecture 1.1 is roughly half a century old, but still unresolved - despite claims of the contrary which appear in the form of preprints or even published papers on a regular basis. We will settle a number of open cases of Conjecture 1.1 in this paper. The following is a classical result [17]. Result 1.2 (Turyn) If an Hadamard matrix of order v exists, then v = 4u 2 for some odd integer u which is not a prime power. v if A sequence a 1,..., a v, a i = ±1, is called a Barker sequence of length v j a i a i+j 1 for j = 1,..., v 1. i=1 The following is well known, see [2, Chapter VI, 14]. Result 1.3 If a Barker sequence of length l > 13 exists, then there is a circulant Hadamard matrix of order l. So, naturally there is also the following. Conjecture 1.4 There are no Barker sequences of length exceeding 13. 2

3 Storer and Turyn [16] proved that there is no Barker sequence of odd length exceeding 13, but the case of even length is still open despite powerful partial results [5, 6, 13, 14, 17]. Curiously, there is only one known open case of Conjecture 1.4: l = 4u 2 with u = , see [8]. This case will be ruled out by our results on circulant Hadamard matrices. The most powerful known results on Conjectures 1.1 and 1.4 were obtained by the so-called field descent method, see [6, 13, 14]. It is this method we will improve upon in the present paper. To make full use of our improvements, we will combine them with results of Turyn [17], McFarland [10], and Chan [4]. 2 Preliminaries To state and prove our results on Conjecture 1.1, we will use the language of groups rings and difference sets. We first fix some notation. Let G be a finite group. We will always identify a subset A of G with the element g A g of the integral group ring Z[G]. For B = g G b gg Z[G] we write B ( 1) := g G b gg 1 and B := g G b g. A group homomorphism G H is always assumed to be extended to a homomorphism Z[G] Z[H] by linearity. For convenience, we write ζ m = e 2πi/m for any integer m. A (v, k, λ, n)-difference set in a finite group G of order v is a k-subset D of G such that every element g 1 of G has exactly λ representations g = d 1 d 1 2 with d 1, d 2 D. The positive integer n = k λ is called the order of the difference set. For an introduction to difference sets, see [2, Chapter VI]. In the group ring language, difference sets can be characterized as follows [2, Chapter VI, Lemma 3.2]. Lemma 2.1 Let D be a k-subset of a group G of order v. Then D is a (v, k, λ, n) difference set in G if and only if in the group ring Z[G] the following equation holds: DD ( 1) = n1 G + λg. 3

4 In this paper, we mainly deal with Hadamard difference sets, i.e., difference sets with parameters (v, k, λ, n) = (4u 2, 2u 2 u, u 2 u, u 2 ) where u is a positive integer. The following is well known, see [2, Chapter VI, 14]. Result 2.2 If a circulant Hadamard matrix of order 4u 2 exists, then there is an Hadamard difference set in the cyclic group of order 4u 2. Next, we state a result concerning Hadamard difference sets, which essentially is a consequence of [10, Thm. 3.1]. Result 2.3 (McFarland) Let G be an abelian group of order 4u 2 w 2 where u and w are odd and coprime, and assume the existence of an Hadamard difference set D in G. Let H be the subgroup of G of order 4u 2. If χ(d) 0 (mod w) for all characters χ of G of order dividing 4u 2, then there is an Hadamard difference set in H. For convenience, we recall the following definition. Definition 2.4 Let p be a prime, let m be a positive integer, and write m = p a m with (p, m ) = 1, a 0. If there is an integer j with p j 1 ( mod m ), then p is called self-conjugate modulo m. A composite integer n is called selfconjugate modulo m if every prime divisor of n has this property. As we shall see, we need to deal with congruences of the form XX 0 (mod u). The first result in this direction is due to Turyn [17] and based on self-conjugacy. Result 2.5 [17] Assume that A Z[ζ m ] satisfies AA 0 (mod t 2b ) where b, t are positive integers, and t is self-conjugate modulo m. Then A 0 (mod t b ). 4

5 The next result concerns solving equation of the form XX = u 2 in Z[ζ p ] without using self-conjugacy. It is based on a method of Chan [4, Lemmas 2.3, 2.4]. Result 2.6 [14, Thm ] Let X Z[ζ p ] be a solution of XX = u 2 where p is an odd prime with (u, p) = 1 and u is a positive integer. Write u = s i=1 qa i i where the q i s are distinct primes. If gcd(ord p (q 1 ),..., ord p (q s )) > 2u(p 1), p then X 0 (mod u). The following two results are essential tools for the methods in the present paper. The first one is due to Kronecker. See [3, Section 2.3, Thm. 2] for a proof. Result 2.7 [Kronecker] An algebraic integer all of whose conjugates have absolute value 1 is a root of unity. Note that Result 2.7 implies that any cyclotomic integer of absolute value 1 must be a root of unity since the Galois group of a cyclotomic field is abelian. A proof of the following result can be found in [14, Thm ], for instance. Result 2.8 Let p be a prime, and let m be a positive integer. Let P be a prime ideal above p in Z[ζ m ], and write m = p a m with (m, p) = 1. If σ Gal(Q(ζ m )/Q) satisfies σ(ζ m ) = ζ pj m σ(p ) = P. for some positive integer j, then For later application, we record the following, which is a consequence of [13, Lem. 2.5]. Result 2.9 Let m > 1 be an integer and X Z[ζ m ] such that X = m 1 a iζ i m with 0 a i C for some constant C. If X 0 (mod u) for some integer u, then u 2 s 1 C where s is the number of distinct prime divisors of m. 5

6 For a prime p and a positive integer t, let ν p (t) be defined by p νp(t) t, i.e. p νp(t) is the highest power of p dividing t. By D(t) we denote the set of prime divisors of t. The following definition is required for the application of the field descent method [13]. Definition 2.10 Let m, n be integers greater than 1. For q D(n) let { p D(m)\{q} p if m is odd or q = 2, m q := 4 p D(m)\{2,q} p otherwise. Set { b(2, m, n) = max ν2 (q 2 1) + ν 2 (ord mq (q)) 1 } and q D(n)\{2} { b(r, m, n) = max νr (q r 1 1) + ν r (ord mq (q)) } q D(n)\{r} for primes r > 2 with the convention that b(2, m, n) = 2 if D(n) = {2} and b(r, m, n) = 1 if D(n) = {r}. We define F (m, n) := gcd(m, p D(m) The following result was proved in [13]. p b(p,m,n) ). Result 2.11 Assume XX = n for X Z[ζ m ] where n and m are positive integers. Then for some j. Xζ j m Z[ζ F (m,n) ] The next result is [14, Thm ]. Result 2.12 (F-bound) Let X Z[ζ m ] be of the form X = m 1 a i ζ i m with 0 a i C for some constant C and assume that n := XX is an integer. Then n C2 F (m, n) 2 4ϕ(F (m, n)). 6

7 3 Results Theorem 3.1 Assume the existence of a (v, k, λ, n) difference set in a cyclic group G. Let m be a positive integer such that m 2 divides n, and let w be a divisor of v such that m is self-conjugate modulo v/w. Then n w2 F (v/w, n/m 2 ) 2 4ϕ(F (v/w, n/m 2 )). Proof Let χ be character of G of order v/w. Then χ(d) 0 (mod m) by Result 2.5 since m is self-conjugate modulo v/w. Write X = χ(d)/m. Then X 2 = n/m 2 and thus Xζ j v/w Z[ζ F ] for some integer j by Result 2.12 where F = F (v/w, n/m 2 ). Hence Z = χ(d)ζ j v/w = mxζj v/w Z[ζ F ], too. Note that Z = v/w 1 a i ζ i v/w with 0 a i w since χ has order v/w. Note that Z 2 = χ(d) 2 = n. Hence the theorem follows by applying Theorem 2.12 to Z. Example 3.2 Let u = = This is the seventh smallest value u such that the existence of a circulant Hadamard matrix of order 4u 2 is still open, see [8]. We take v = 4u 2, n = u 2, w = and m = 11 in Theorem 3.1. Then F (v/w, n/m 2 ) = and get ( ) 2 (2 232 ) 2 ( ) 2, a contradiction. Thus no circulant Hadamard matrix of order 4u 2 exists for u = Remark 3.3 According to Mossinghoff [8], there are 1576 positive integers u for which the existence of a circulant Hadamard matrix of order 4u 2 is still open. Theorem 3.1 rules out 135 of these cases. The details are provided in [15]. For each case, the triple [u, w, m] is given which is needed for the application of Theorem 3.1 to rule out the existence of a circulant Hadamard matrix of order 4u 2. Lemma 3.4 Let p be an odd prime and let a and u be positive integers. Suppose X Z[ζ p a] satisfies X 2 = u 2. Let w be a divisor of u which is self-conjugate modulo p. Let q 1,...,q k be the prime divisors of u/w. If f := gcd(ord p (q 1 ),..., ord p (q k )) > u2 w 2, 7

8 then X 0 (mod u). Proof If f is even, then u is self-conjugate modulo p which implies X 0 (mod u) by Result 2.5. Hence we can assume that f is odd. Since w is self-conjugate modulo p, we can write X = wy for some Y Z[ζ p a] by Lemma 2.5. If w = u, there is nothing to show, so we can assume w < u. Note Y 2 = u 2 /w 2. Let t be an integer with ord p a(t) = f. Note that t 1 0 (mod p) since f > 1. Define σ Gal(Q(ζ p a)/q) by σ(ζ p a) = ζp t a. Note that t ( mod p a ) for some α i Z for all i by the definition of f. Hence, by Result q α i i 2.8, all prime ideals above (Y ) in Z[ζ p a] are invariant under σ. Hence σ(y ) = δκy (1) where δ = ±1 and κ is some p a th root of unity. Since t 1 0 (mod p), there is an integer r with r(t 1) 1 (modp a ). Write Z = Y κ r. Then σ(z) = δκ 1+rt Y = δκ r Y = δz. Applying σ repeatedly to this equation, we get Z = σ f (Z) = δ f Z. Thus δ = 1 as f is odd. Hence σ(z) = Z. Write p a 1 Z = ζp i az i (2) with Z i Z[ζ p ]. Since σ(z) = Z, we have p a 1 1 ζ i p az i = p a 1 1 ζ it p aσ(z i). (3) Note that σ(z i ) Z[ζ p ] and that {1, ζ p a,..., ζ pa 1 1 p } is independent over a Z[ζ p ]. Hence (3) shows that Z i 0 implies Z it 0 for all i where the indices are taken modulo p a 1. Note that, by the definition of t, all orbits {0} of multiplication by t on Z/p a 1 Z have length f. Hence, if there is i > 0 with Z i 0, then there are at least f indices j with Z j 0. 8

9 Recall that Z 2 = Y 2 = u 2 /w 2. Again, since {1, ζ p a,..., ζ pa 1 1 p a } is independent over Z[ζ p ], this implies p a 1 1 Z i Z i = u2 w 2. (4) Now assume Z i 0, and let Γ = Gal(Q(ζ p /Q). Since τ Γ τ(z iz i ) = N Q(ζp)/Q(Z i Z i ) 1, we have τ(z i Z i ) p 1 (5) τ Γ by the inequality of arithmetic and geometric means. number of indices i with Z i 0. Then Now let N be the N(p 1) τ Γ p a 1 1 Z i Z i = u2 (p 1) w 2 by (4) and (5). Thus N u 2 /w 2. Above we have seen that N f if Z i 0 for some i > 0. Since f > u 2 /w 2 by assumption, we infer Z i = 0 for all i > 0. Hence Z = Z 0 Z[ζ p ]. Note f > u2 w 2 2u w > 2(p 1)u pw (6) since u/w 2. Because of Z 2 = u 2 /w 2 and (6), Result 2.6 implies Z 0 (mod u/w). Hence X = wy = κ r wz 0 (mod u). Theorem 3.5 Let p be an odd prime, let a, m be positive integers with (m, p) = 1, and write u = p a m. Let r be any divisor of m which is selfconjugate modulo p, and let q 1,..., q s be the prime divisors of m/r. If f := gcd(ord p (q 1 ),..., ord p (q k )) > then there is no circulant Hadamard matrix of order 4u 2. u2 p 2a r 2 and pa > 2m, (7) Proof Suppose D is a Hadamard difference set in a cyclic group G corresponding to a circulant Hadamard matrix of order 4u 2. Let χ be a character of G of order p 2a or 2p 2a. Then χ(d) Z[ζ p 2a] and χ(d) 2 = u 2 by (2.1). 9

10 Since w := p a r is self-conjugate modulo p by assumption, Lemma 3.4 implies χ(d) 0 (mod u), i.e., χ(d) = ±ηu for some p 2a th root of unity η. Let W be the subgroup of order 2u 2 of G, and write D = A + Bg where A, B W, and g is an element of order 4 in G. Let χ 1 be a character of G of order p 2a and let χ 2 be the unique character of G defined by χ 2 (h) = χ 1 (h) for all h W and χ 2 (g) = 1. Note that the order of χ 2 is 2p 2a. By what we have shown above, χ i (D) = δ i η i u for some p 2a th roots of unity η i and δ i { 1, 1}, i = 1, 2. Replacing D by Dk for some k G, if necessary, we can assume χ 2 (D) = u. Note that χ 1 (D) = χ 1 (A) + χ 1 (B) and χ 2 (D) = χ 1 (A) χ 1 (B). Hence χ 1 (D) + χ 2 (D) = 2χ 1 (A) and thus δ 1 η 1 u + u 0 (mod 2). Since u is odd, this implies δ 1 η (mod 2) and thus η 1 = 1. Hence we have χ 2 (D) = χ 1 (D) or χ 2 (D) = χ 1 (D). If χ 2 (D) = χ 1 (D), then 2χ 1 (A)χ 1 (D)+ χ 2 (D) = 2u, and if χ 2 (D) = χ 1 (D), then 2χ 1 (B) = χ 1 (D) χ 2 (D) = 2u. In summary, χ 1 (A) = u or χ 1 (B) = u. Since A, B W the kernel of χ 1 on W has order 2u 2 /p 2a = 2m 2, we can write χ 1 (A) = p 2a 1 a i ζ p 2a with 0 a i 2m 2, and χ 1 (B) has a similar representation. Hence u 2m 2 by Result 2.9, i.e., p a 2m, contradicting the assumptions. Example 3.6 Let u = = This corresponds to the only known open case of the Barker sequence conjecture, see [8]. Assume a circulant Hadamard matrix of order 4u 2 exists. We take p = , a = 1, m = , r = 2953 in Theorem 3.5. We find f = gcd(ord p (13), ord p (41)) = > = p 2a r 2 and p = > = 2m. Hence (7) is satisfied and we get a contradiction. Hence no circulant Hadamard matrix of order 4u 2 exists. Remark 3.7 Theorem 3.5 rules out 22 of the open cases of circulant Hadamard matrices of order 4u 2 with u in Mossinghoff s list [8]. The details are provided in [15]. For each case, the quadruple [u, p, m, r] is given which is needed for the application of Theorem u2

11 Corollary 3.8 No Barker sequence of length L exists with 13 < L Proof If there is a Barker sequence of length L with 13 < L , then L = 4u 2 with u = by the results in [5, 6, 17] and a computer search described in [8]. But a Barker sequence of this length cannot exist by Example 3.6 since the existence of a Barker sequence of even length L implies the existence of a circulant Hadamard matrix of order L. Lemma 3.9 Let p 3 (mod 4) be a prime, and let u be an integer all of whose prime divisors are 3 (mod 4). Let X Z[ζ 4p a] with X 2 = u 2. Then Xζ j 4 Z[ζ p a] for some integer j. Proof Suppose q is a prime divisor of u such that t := ord p (q) is even. Then t 2 (mod 4) as p 3 (mod 4). Hence q t/2 1 (mod p) and thus q pa 1t/2 1 (mod p a ). Moreover q pa 1t/2 1 (mod 4) since q 3 (mod 4) and p a 1 t/2 is odd. We conclude q pa 1t/2 1 (mod 4p a ), i.e., q is self-conjugate modulo 4p a. Using Result 2.5, we conclude X 0 ( mod q). Repeating this argument, if necessary, we can write X = wy with Y 2 = u 2 /w 2 where w is a positive integer dividing u, and all prime divisors of u/w are of odd order modulo p. Now let q be a prime divisor of u/w, so s := ord p (q) is odd. Then q tpa 1 1 (mod p a ) and q tpa 1 3 (mod 4) since q 3 (mod 4) and tp a 1 is odd. Hence, by Result 2.8, the automorphism of Q(ζ 4p a) defined by σ(ζ p a) = ζ p a and σ(ζ 4 ) = ζ 4 fixes all prime ideals of Z[ζ 4p a] above q. Since this holds for all prime divisors of u/w and Y 2 = u 2 /w 2, we conclude that σ fixes the ideal Y Z[ζ 4p a]. Hence σ(y ) = δy for some unit δ of Z[ζ 4p a] with δ = 1. By Result 2.7, we have δ = ζ j p aik for some integers j, k, i.e., σ(y ) = ζ j p aζk 4 Y. (8) 11

12 Applying σ to (8), we get Y = σ 2 (Y ) = ζ j p a( ζ 4) k σ(y ) = ζ 2j p ay. This implies ζ j pa = 1, i.e., σ(y ) = ζ k 4 Y. (9) Now write Y = A + Bζ 4 with A, B Z[ζ p a]. Then (9) implies A Bζ 4 = ζ k 4 (A + Bζ 4 ). (10) If k = 0, then B = 0 by (10). If k = 1, then A = B by (10) and thus Y = A(1 ζ 4 ). But then Y 2 = 2 A 2, contradicting the assumption that u is odd. The cases k = 2, 3 are treated similarly. In summary, we have A = 0 or B = 0 in all cases. This concludes the proof. Theorem 3.10 Let u be an integer all of whose prime divisors are congruent to 3 modulo 4 and let p be one these prime divisors. Let w be a divisor of u which is self-conjugate modulo p. Let q 1,...,q k be the prime divisors p of u/w. If gcd(ord p (q 1 ),..., ord p (q k )) > u2 w 2, (11) then no circulant Hadamard matrix of order 4u 2 exists. Proof Suppose D is a Hadamard difference set in a cyclic group G corresponding to a circulant Hadamard matrix of order 4u 2. If χ is a character of G of order dividing 4, then χ(d) 0 (mod u) by Result 2.5 since u is self-conjugate modulo 4. Write u = p a m with (p, m) = 1. Let χ be a character of G of order 2 b p c where 0 b 2, 1 c 2a. Then χ(d) Z[ζ 4p c] and χ(d) = u 2 by Result 2.1. Lemma 3.9 shows that χ(d)η Z[ζ p c]. In view of (11), Lemma 3.4 implies χ(d) 0 (mod u). In summary, we have shown χ(d) 0 ( mod u) for all characters χ of G of order dividing 4p 2a. Thus, by McFarland s Result 2.3, there is a Hadamard difference set in the cyclic group of order 4p 2a. Result This contradicts Turyn s

13 Example 3.11 Let u = = which corresponds to the second smallest open case of circulant Hadamard matrices, see [8]. We take p = 71, w = in Theorem Then u/w = 3 and ord p (3) = 35 > 9 = u 2 /w 2. Hence no circulant Hadamard matrix of order 4u 2 exists by Theorem Remark 3.12 Theorem 3.10 rules out 63 cases of Mossinghoff s list [8] of open cases of the circulant Hadamard matrix conjecture. Please see [15] for the details. For each case, the triple [u, p, r] is given where p and r are the numbers needed for the application of Theorem Acknowledgement Part of this research was done during a visit of the second author at the Unversität Augsburg who thanks Dieter Jungnickel for his kind hospitality. References [1] K.T. Arasu, S.L. Ma: Abelian difference sets without self-conjugacy. Des. Codes Cryptogr. 15 (1998), [2] T. Beth, D. Jungnickel, H. Lenz: Design Theory (2nd edition). Cambridge University Press [3] Z.I. Borevich, I.R. Shafarevich: Number Theory. Academic Press, New York/San Francisco/London (1966). [4] W.K. Chan: Necessary Conditions for Menon Difference Sets. Designs, Codes and Cryptography 3 (1993), [5] S. Eliahou, M. Kervaire, B. Saffari: A new restriction on the length of Golay complementary sequences. J. Comb. Theory Ser. A 55 (1990), [6] K.H. Leung and B. Schmidt, The Field Descent Method. Des. Codes Crypt. 36 (2005), [7] S.L. Ma: Polynomial addition sets. Ph.D. thesis, University of Hong Kong,

14 [8] M. Mossinghoff: Wieferich pairs and Barker sequences, Des. Codes Cryptogr. 53 (2009), [9] R.L. McFarland: On multipliers of abelian difference sets. Ph.D. Dissertation, Ohio State University [10] R.L. McFarland: Sub-difference sets of Hadamard difference sets. J. Comb. Theory A 54 (1990), [11] A. Pott: Finite geometry and character theory. Lecture Notes 1601, Springer [12] H.J. Ryser: Combinatorial Mathematics. Wiley, New York [13] B. Schmidt: Cyclotomic integers and finite geometry. J. Am. Math. Soc. 12 (1999), [14] B. Schmidt: Characters and cyclotomic fields in finite geometry. Lecture Notes in Mathematics 1797, Springer [15] B. Schmidt: Data on Circulant Hadamard Matrices. [16] J. Storer, R. Turyn: On binary sequences. Proc. Amer. Math. Soc. 12 (1961), [17] R.J. Turyn: Character sums and difference sets. Pacific J. Math. 15 (1965),